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1) Given a measure space, $f$ a non-negative measurable function and $A$ in the $\sigma$-algebra such that $\mu(A)=0$, prove that $\displaystyle\int_{A} f\;d\mu=0$ .

My try:

$$0\leq \int_{A}f\;d\mu=\int fI_{A}\;d\mu\leq \int \sup{f(A)}I_{A}\;d\mu=\sup{f(A)}\int I_{A}\;d\mu=\sup{f(A)}\mu(A)=0\;,$$ then $\displaystyle\int_{A}f\;d\mu=0\,.$

2) Given a measure space, let $f$ be an integrable function. Show that $\displaystyle \int_{\{|f|>t\}} |f|\;d\mu\rightarrow 0$ as $t\rightarrow\infty$ .

My try: since $f$ is integrable, $|f|\lt \infty$ almost everywhere, $\mu (\{|f|>t\}) = \mu(\Omega)- \mu(\{|f|\leq t\})$, and as $t\rightarrow\infty$, $\mu (\{|f|>t\})<\epsilon$, then $$\int_{\{|f|>t\}} |f|\;d\mu=\int |f|I_{\{|f|>t\}}\;d\mu\leq\max\{|f|\}\int I_{\{|f|>t\}}=\max\{|f|\}\;\mu(\{|f|>t\})<\max\{|f|\}\;\epsilon\;.$$ Then it's done; I can take $\epsilon=\frac1t$.

Well, both demonstrations seem right for me. I will be thankful if someone tells me what's wrong on them.

Greetings.

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The first argument looks good to me (at least if we assume that $f\ge0$; maybe you want to apply your argument to $|f|$). The second part doesn't seem so good. For example, what happens if $\mu(\Omega)=\infty$? And why is $\max\{|f|\}$ finite? –  mac May 17 '11 at 17:35
    
Yes, you're right mac f should be ≥0 and well in the second it's true, what if $\mu(\Omega)=\infty$? there i don't know how to heal it but your second observation can be easily heal by taking, let's say, the associated function of f such that it takes the value of any finite number (1 for example) where f takes $\infty$ as this set is of measure zero the intregrals of both functions coincide. –  Ivan3.14 May 17 '11 at 18:24
    
I don't think your proposal to "heal" the possibility that $\max{|f|}=\infty$ will work, since if $f$ is any unbounded real-valued function, such as $f(x)=e^x$, then you'll be forced to interpret $\max{|f|}$ (really $\sup |f|$, I guess) as $\infty$. –  mac May 17 '11 at 19:40
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3 Answers

up vote 5 down vote accepted

Edit: I missed some things in my first answer and updated my answer accordingly.

Edit 2: Not only that I missed some things in my first answer, I missed the point completely. I leave the answer here for instructional purposes. Tasadduk's answer is definitely the way to go about 2).


I agree with mac that the first solution is almost ok, but some extra care has to be taken (the supremum might be infinite and even if one could argue that $\infty \cdot 0 = 0$ in measure theory, a clear cut argument is certainly preferable to a convention).

I would first prove that for a non-negative and bounded functions $f$ we have $\int_{A} f \,d\mu = 0$, for this you can use your argument. The monotone convergence theorem then implies that it holds for all non-negative measurable functions, simply by approximating $f$ by $f_{n} = \min{\{f,n\}}$ from below.

Remark. The equality $\int_{A} f = 0$ holds true for all measurable $f$, because a general $f$ can be written as $f = f_+ - f_-$ with $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$ and use the definition $\int_{A} f\,d\mu = \int_{A} f_+ \,d\mu - \int_{A} f_-\,d\mu$ provided at least one of the integrals of the right hand side is finite.


As mac also pointed out, there are quite a few problems in the second proof and I don't see how to save it using your idea.

Here's how I would do it. Let $A_{t} = \{|f| \gt t\}$ and note that for $s \lt t$ we have $A_{s} \supset A_t$. Define $$a_{t} = \int_{\{|f| \gt t\}} |f|\,d\mu = \int_{A_t} |f|\,d\mu.$$ Since $|f| \geq 0$ and $A_{s} \supset A_{t}$ we have $a_{s} \geq a_{t}$ for $s \leq t$. We want to show that $a_{t} \to 0$ as $t \to \infty$. By monotonicity of $t \mapsto a_{t}$, it suffices to show that $a_{n} \to 0$ as $n \to \infty$ runs through the natural numbers. Since $a_{n} \geq 0$ and $a_{n}$ is monotonically decreasing, the limit $a = \lim_{n \to \infty} a_{n}$ exists, and we want to show that $$ a = 0.$$

Now let $f_{n}(x) = \min{\{|f(x)|,n\}}$ and note that $f_{n} \to |f|$ pointwise (and monotonically). By the monotone convergence theorem (or the dominated convergence theorem, applicable since $f_{n} \leq |f|$ and $|f|$ is integrable, if you prefer) we have

$$ \int |f|\,d\mu = \lim_{n \to \infty} \int f_{n}\,d\mu.$$

On the other hand $n \leq |f|$ on $A_{n}$ and $0 \leq f_{n} \leq |f|$ on $\Omega$ imply $$ \int f_{n}\,d\mu = \int_{A_{n}} n \,d\mu + \int_{\Omega \smallsetminus A_{n}} f_{n}\,d\mu \leq \int_{A_{n}} |f|\, d\mu + \int |f| \,d\mu = a_{n} + \int |f|\,d\mu.$$

Passing to the limit on both sides of the estimate $\int f_{n} \leq a_{n} + \int |f|$ we get $$\int |f|\,d\mu = a + \int |f|\,d\mu$$ and as $\int |f|\,d\mu \lt \infty$ we conclude $a = 0$, as we wanted.

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1) I think a more natural argument would be to first prove the result in the case where $f$ is a simple function, then extend via the monotone convergence theorem. That being said, I like your solution, although something about it seems weird.

2) This argument fails if $\max|f|$ doesn't exists; i.e. if $\sup|f| = \infty$. I recommend proving by contradiction: If the result fails, then there is $\epsilon>0$ and a sequence $t_1 < t_2 < t_3 <\cdots$ with $t_i \to \infty$ such that $$\int_{\{|f|>t_{i}\}} |f| \, du > \epsilon$$ for all $i$. Using the monotone convergence theorem, you can then show that $$\sum^{\infty}_{i=k} \int_{\{t_i < |f| \leq t_{i+1}\}} |f|\, du = \int_{\{|f|>t_k\}}|f|\, du >\epsilon$$ for every $k$, so that sum on the left-hand side must be infinite (for each $k$), contrary to $f$'s integrability.

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I think the usual measure theory convention is that $\infty \times 0=0$. So I think the first argument should work even if $\sup|f(A)|=\infty$. –  mac May 17 '11 at 19:37
    
You're right, I missed your remark on the first part and having an argument is certainly preferable to relying on a convention. I updated my answer accordingly. However, you don't have to go all the way back to simple functions. The argument in the question goes through for bounded non-negative functions. On the other hand, I prefer my direct argument to your second one because spelling out the details both come down to approximately the same length. –  t.b. May 17 '11 at 19:38
    
@mac But you can't factor $\infty$ out of an integral, only finite constants. –  Quinn Culver May 17 '11 at 19:41
    
@Theo Notice that in the first part the asker stipulated that $f$ is nonnegative. –  Quinn Culver May 17 '11 at 19:43
    
@Quinn: isn't it true that $\int \infty f\,d\mu = \infty\int f\,d\mu$ for $f\ge0$? –  mac May 17 '11 at 19:47
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Solution to (2): I replaced t with n cause that's how I like it, haha. Let $E_n = \{ x: |f(x)|>n \}$. Note that $E_n \supset E_{n+1}$, we have a decreasing sequence of sets. And $\bigcap_n E_n = \{ x: |f(x)| = \infty \}$. Now since $f$ is integrable, so is $|f|$. We define a measure $\lambda(E) = \int_E |f| d\mu$ for all $E$ in the $\sigma$-algebra. Note that $\lambda(E_1) = \int_{E_1} 1 d\mu\ \leq \int_{E_1} |f| d\mu\ < \infty$. And since $\mu(\bigcap_n E_n) = 0, \lambda(\bigcap_n E_n) = 0$. Thus \begin{equation*} \lim_{n \to \infty} \int_{E_n} |f| d\mu\ = \lim_{n \to \infty} \lambda(E_n) = \lambda(\bigcap_n E_n) = 0. \end{equation*}

Remark: Since $f$ is integrable, so is $|f|$. Therefore if it so happens that $|f| = \infty$ at some set, that set has measure zero! So for this problem, we need not worry about it.

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Whoa! How could I possibly have missed that? Well, nice answer anyway :) I fixed some small typos. –  t.b. May 20 '11 at 5:55
    
Wow... glad you guys liked my answer! This was a good problem, by the way. –  Tasadduk Chowdhury May 20 '11 at 6:24
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